我在r。
中对代码有问题我有一个带有4列和超过600k观察的数据集(问题),其中一列命名为" V3"。本专栏有"什么一天?"之类的问题。我有第二列的第二个数据集(VOC),其中一个列名称为" word"和其他列名"同义词"。如果在我的第一个数据集(问题)中存在从"同义词"列中的第二个数据集(VOC)中存在的,则我想从'Word''列中替换Word。
questions = cbind(V3=c("What is the day today?","Tom has brown eyes"))
questions <- data.frame(questions)
V3
1 what is the day today?
2 Tom has brown eyes
voc = cbind(word=c("weather", "a","blue"),synonyms=c("day", "the", "brown"))
voc <- data.frame(voc)
word synonyms
1 weather day
2 a the
3 blue brown
Desired output
V3 V5
1 what is the day today? what is a weather today?
2 Tom has brown eyes Tom has blue eyes
我编写了简单的代码,但不起作用。
for (k in 1:nrow(question))
{
for (i in 1:nrow(voc))
{
question$V5<- gsub(do.call(rbind,strsplit(question$V3[k]," "))[which (do.call(rbind,strsplit(question$V3[k]," "))== voc[i,2])], voc[i,1], question$V3)
}
}
也许有人会尝试帮助我?:)
我写了第二个代码,但它也不起作用。
for( i in 1:nrow(questions))
{
for( j in 1:nrow(voc))
{
if (grepl(voc[j,k],do.call(rbind,strsplit(questions[i,]," "))) == TRUE)
{
new=matrix(gsub(do.call(rbind,strsplit(questions[i,]," "))[which(do.call(rbind,strsplit(questions[i,]," "))== voc[j,2])], voc[j,1], questions[i,]))
questions[i,]=new
}
}
questions = cbind(questions,c(new))
}
首先,在程序级别或数据导入期间使用stringsAsFactors = FALSE
选项很重要。这是因为r默认值将字符串变成因素,除非您另行指定。因素在建模方面很有用,但是您想对文本本身进行分析,因此您应该确保您的文本不会被强制为因素。
我接近这一点的方式是编写一个可以将每个字符串"爆炸"到向量的函数,然后使用匹配来替换单词。向量再次重新组合到字符串中。
我不确定这将如何获得您的600k记录。您可能会查看一些处理字符串(例如stringr
或stringi
)的R软件包,因为它们可能具有执行其中一些功能。match
的速度往往可以,但是%in%
可以是真正的野兽,具体取决于字符串的长度和其他因素。
# Start with options to make sure strings are represented correctly
# The rest is your original code (mildly tidied to my own standard)
options(stringsAsFactors = FALSE)
questions <- cbind(V3 = c("What is the day today?","Tom has brown eyes"))
questions <- data.frame(questions)
voc <- cbind(word = c("weather","a","blue"),
synonyms = c("day","the","brown"))
voc <- data.frame(voc)
# This function takes:
# - an input string
# - a vector of words to replace
# - a vector of the words to use as replacements
# It returns a list of the original input and the changed version
uFunc_FindAndReplace <- function(input_string,words_to_repl,repl_words) {
# Start by breaking the input string into a vector
# Note that we use [[1]] to get first list element of strsplit output
# Obviously this relies on breaking sentences by spacing
orig_words <- strsplit(x = input_string,split = " ")[[1]]
# If we find at least one of the words to replace in the original words, proceed
if(sum(orig_words %in% words_to_repl) > 0) {
# The right side selects the elements of orig_words that match words to be replaced
# The left side uses match to find the numeric index of those replacements within the words_to_repl vector
# This numeric vector is used to select the values from repl_words
# These then replace the values in orig_words
orig_words[orig_words %in% words_to_repl] <- repl_words[match(x = orig_words,table = words_to_repl,nomatch = 0)]
# We rebuild the sentence again, and return a list with original and new version
new_sent <- paste(orig_words,collapse = " ")
return(list(original = input_string,new = new_sent))
} else {
# Otherwise we return the original version since no changes are needed
return(list(original = input_string,new = input_string))
}
}
# Using do.call and rbind.data.frame, we can collapse the output of a lapply()
do.call(what = rbind.data.frame,
args = lapply(X = questions$V3,
FUN = uFunc_FindAndReplace,
words_to_repl = voc$synonyms,
repl_words = voc$word))
>
original new
1 What is the day today? What is a weather today?
2 Tom has brown eyes Tom has blue eyes